Discussion on: We don't need a ternary operator

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Replies for: That's correct, ifElse accepts functions. So those functions are only executed when the condition is met.

Both expensive gets will run before the condition is even applied to true.

This is still false. Only one will run. Never both.

const isEven = n => n % === 0
const logEven = n => console.log(`${n} is Even!`)
const logOdd = n => console.log(`${n} is Odd`)

ifElse (isEven) (logEven) (logOdd) (10)

//=> "10 is Even!"

You can see logOdd is never called.

Mihail Malo • Sep 9 '18

In your first example

const expensiveGetPlus = x => expensiveGet() + x

ifElse (condition) (expensiveGetPlus (1)) (expensiveGetPlus (-1)) (true)

both run.

const expensiveGet = () => {
  console.log("doing expensive get")
  return 2
}

JavaScript Joel • Sep 9 '18

Yes you are correct. The expensiveGet method needs to take 2 arguments for it to work the way it is being called in ifElse.

I have created a working example that you can run.

const S = require('sanctuary')

const expensiveGet = (y) => {
  console.log("doing expensive get for", y)
  return 2
}
const condition = x => x === 100
const expensiveGetPlus = x => y =>
  expensiveGet(y) + x

S.ifElse (condition) (expensiveGetPlus (1))
  (expensiveGetPlus (-1)) (100)
//=> "doing expensive get for 100"

This example will show that ifElse only executes one of the functions.